Thank you The distribution of incomes of adults in the U.S. does not follow the

Question

Thank you The distribution of incomes of adults in the U.S. does not follow the normal curve. A researcher took a random sample of 400 adults in the U.S. The average income in the sample was $48, 760 and the SD of the sample incomes was $18, 518. (a) Find, if possible, a 92%-confidence interval for the average income of all adults in the U.S. (b) True or false, and explain briefly. i. Since the population does not follow the normal curve, you can't use the curve to get confidence intervals. ii. About 92% of all the adults in the U.S. have incomes lying within the 92%-confidence interval. iii. There is a 92% chance that the average of all adults lies in the 92% confidence interval. iv. Since this is a well-designed sample survey, the sample average is very likely to equal the population average. You send a yardstick to a local laboratory for calibration, asking that the procedure be repeated twenty-five times. They report the following values: 35.95 inches (8 times), 36 inches (9 times), 36.05 inches (8 times) (a) If you send the yardstick back for a twenty-sixth calibration, you would expect to get 36 inches give or take- .04 inches or so .05 inches or so .10 inches or so (Say whether each of (b-d) is true or false and explain your answers briefly.) (b) The average of all 25 measurements is off 36 inches by .008 inches or so. (c) A 95%-confidence interval for the length of the yardstick is 36 plusminus .016 inches. (d) About 95% of the measurements were in the range 36 plusminus .016 inches. (For this entire problem, you may assume the Gauss model, with Do bias.)

Explanation / Answer

Let the given random sample of size n = 400 adults in US have average income x = $48,700, and standard deviation S.D. = $18,518

It is also given that distribution of income of adults in US does not follow the normal curve, but by using Central Limit Theorem which is a statistical theory that states that given a sufficiently large sample size from a population with a finite level of variance, the mean of all samples from the same population will be approximately equal to the mean of the population,

So which also means that the mean of any distribution is always follows normal distribution, with mean x and standard error S.E. = S.E. / sqrt( n )

Let x ~ N( x, S.E. / sqrt( n ) ) = N ( 48,700, 18518/sqrt(400) ) = N ( 48700, 925.9 )

Now let the 92% confidence interval for the average income of all adults in the U.S. is given by,

( x - Z/2 * S.E. ,   x + Z/2 * S.E. )

Let = 1 - c = 1 - 0.92 = 0.08, hence from standard normal probability table we can find Z/2 for /2 = 0.04 as

Z/2 = 1.75, Now let plug the values in above formulas we will get, ( 48700 - 1.75 * 925.5, 48700 + 1.75 * 925.5 )

so the 92% confidence interval for income of US adults is given by ( 47079.68, 50320.33)

(b) True or False,

(i) False, We have sufficient sample size hence we can approximate it to C.L.T. Hence we can get confidence interval by using approximated normal curve.

(ii) False, Instead we can say that about 92% of all the adults in the U.S. have an average income lying within the 92% confidence interval ( which is calculated above )

(iii) False, 92% chance that the avergae income of all adults lies in the 92% confidence interval

(iv) True. Sample average is very likely to equal to population average by using central limit theorem, with enough sample size.

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